The Winning EQUATION - California State University, …vcmth00m/factors.pdfdivide any whole number a by any counting number b, assuming a > b, special importance is attached to the

© 1999, CISC: Curriculum and Instruction Steering Committee The WINNING EQUATION

CISC - Curriculum & Instruction Steering CommitteeCalifornia County Superintendents Educational Services AssociationPrimary Content Module IV NUMBER SENSE: Factors of Whole Numbers

The Winning EQUATIONA HIGH QUALITY MATHEMATICS PROFESSIONAL DEVELOPMENTPROGRAM FOR TEACHERS IN GRADES 4 THROUGH ALGEBRA II

STRAND: NUMBER SENSE: Factors of Whole Numbers

MODULE TITLE: PRIMARY CONTENT MODULE IV

MODULE INTENTION: The intention of this module is to inform and instruct participants inthe underlying mathematical content in the area of factoring whole numbers.

THIS ENTIRE MODULE MUST BE COVERED IN-DEPTH.The presentation of these Primary Content Modules is a departure from past professionaldevelopment models. The content here, is presented for individual teacher’s depth ofcontent in mathematics. Presentation to students would, in most cases, not address thegeneral case or proof, but focus on presentation with numerical examples.

TIME: 2 hours PARTICIPANT OUTCOMES:

•Demonstrate understanding of factors of whole numbers.•Demonstrate understanding of the principles of prime and composite numbers.•Demonstrate how to determine the greatest common factor and the least commonmultiple when relating two whole numbers.

CISC - Curriculum & Instruction Steering Committee 2

California County Superintendents Educational Services Association

Primary content Module IV NUMBER SENSE: Factors of Whole Numbers


Pre-Post Test

T-1

T-2

T-3

T-4

PRIMARY CONTENT MODULE IVNUMBER SENSE: Factors of Whole Numbers

Facilitator’s Notes

Ask participants to take the pre-test. After reviewing the results ofthe pre-test proceed with the following lesson on factors of wholenumbers.

Divisability: While expressions of the form a = b • q + r enable us todivide any whole number a by any counting number b, assuming a >b, special importance is attached to the case when r = 0. A number b> 0 is called a divisor or factor of a if there exists a whole number qso that a = b • q.

Since 0 = b • 0, 0 is divisible by any b > 0.

Whenever a > 0 and a = b • q, there is a rectangular array of b • q dotsthat corresponds to the factorization of a by b.

Zero and One: Note that,

a = 1 • a or a ÷ 1 = a

and

0 = a • 0 or 0 ÷ a = 0

leads to the conclusion that 0 is divisible by any number a ≠ 0 andany number a is divisible by 1.

Caution: Division by zero is not allowed.

Suppose a ÷ 0 = q Then a = 0 • q.

This is impossible if a ≠ 0. If a = 0, any value of q works, but fordivision there can only be one answer.

An example is a = b • q, assuming a > b,

24 = 4 • 6a factors as b • q24 factors as 4 • 6

Have participants think about how to represent 24 as factors anotherway, and how to represent factors of 7.





T-5

T-6

T-6AT-6B/H-6

T-6C

[Optional:T-20 through T-23]

T-7

Factoring is an important skill for later applications (e.g., it arises inthe addition of fractions) and a concept of interest in its own right.An engaging way to introduce students to this topic is to relate it tothe study of prime numbers.

A whole number is said to be prime if it has exactly two factors: oneand itself. This definition keeps 1 from being a prime. A number > 1that is not prime is called a composite. The first of these definitionsleads to the following list of primes:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,…

Primes have been studied for thousands of years, and many of theirimportant properties were established by the ancient Greeks. Amongthese is the fact that the above list goes on indefinitely; i.e., there isno largest prime.

Show participants short cuts for testing if whole numbers are divisibleby other whole numbers.

Number Shortcut2 ones digit is 0 or even3 sum of digits is divisible by 34 last two digits are divisible by 45 ones digit is 0 or 56 rules for 2 and 3 both work8 last 3 digits are divisible by 89 sum of digits are divisible by 9

10 ones digit is 0

Use appendix section on justification for divisibility rules at thispoint, if desired.

Every whole number > 1 can be written as the product of primefactors.

For example: “Chipaway” technique24 = 2 • 12

= 2 • 2 • 6= 2 • 2 • 2 • 3= 23 • 3

“Split Asunder” Technique24 = 4 • 6

= 2 • 2 • 2 • 3= 23 • 3

“Chipaway” technique1200 = 2 • 600





T-8

T-8A/H-8

T-9

T-10A

T-10B

T-11/H-11

= 2 • 2 • 300= 2 • 2 • 150 = 2 • 2 • 2 • 2 • 75= 2 • 2 • 2 • 2 • 3 • 25= 2 • 2 • 2 • 2 • 3 • 5 • 5= 24 • 3 • 52

“Split Asunder” Technique1200 = 30 • 40

= 5 • 6 • 5 • 8= 5 • 2 • 3 • 5 • 2 • 2 • 2= 24 • 3 • 52

These techniques may be accomplished with the process of factortrees. Transparency T-8 demonstrates this process.

Have participants use both techniques on 60 and 500 and draw thecorresponding factor tree.

The Fundamental Theorem of Arithmetic asserts that except fororder, you will obtain the same list of primes regardless of the methodused to arrive at a prime factorization.

It states that every composite number greater than one can beexpressed as a product of prime numbers. Except for the order inwhich the prime numbers are written, this can only be done in oneway.

Factors and GCFs: Aside from being able to factor whole numbersinto products of primes, it is also important to be able to develop listscontaining all factors (prime and composite) of a particular number.

For example,

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.

The factors of 37 are: 1, 37.

The factors of 64 are: 1, 2, 4, 8, 16, 32, 64.

Except in the case of perfect squares, factors appear in pairs whoseproduct is the number being factored.

Have participants find the factors of 32 and of 80 using H-11. Haveparticipants discuss the difference between “finding all factors” and“finding the prime factorization.” Example: A list of all factors





T-12

T-13AT-13B

T-13C/H-13

of 80 is 1, 2, 4, 5, 8, 10, 16, 20, 40, 80 because 80 is divisible by all ofthese. The prime factorization of 80 is 2 • 2 • 2 • 2 • 5 = 24 • 5because 2 and 5 are both primes.

Given two numbers such as 24 and 64, we can form a list of commonfactors. Referring to the above lists of factors of 24 and 64, weconclude that

The common factors of 24 and 64 are 1, 2, 4, and 8.

The last list leads to the important concept of greatest commonfactor (GCF):

The greatest common factor of 24 and 64 is 8.

This is sometimes written as GCF(24, 64) = 8.

A) List of factors

30: 1, 2, 3, 5, 6, 10, 15, 3096: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Common factors of 30 and 96 are 1, 2, 3, 6GCF(30, 96) is 6.

B) List of factors

90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 9075: 1, 3, 5, 15, 25, 75

Common factors of 90 and 75 are 1, 3, 5, 15GCF(90, 75) is 15.

It is possible to determine GCF(a,b) from the prime factorizations ofa and b. Follow the examples on the slides.

A) 30 = 2 • 3 • 5 = 2 • 3 • 596 = 2 • 2 • 2 • 2 • 2 • 3 = 25 • 3

GCF(30, 96) = 2 • 3 = 6

B) 90 = 2 • 3 • 3 • 5= 2 • 32 • 575 = 3 • 5 • 5 = 3 • 52

GCF(90, 75) = 3 • 5 = 15





T-14A

T-14B

T-14C/H-14

T-15A

T-15B

T-15C/H-15

C) 4500 = 22 • 32 • 53

4050 = 21 • 34 • 52

GCF(4500, 4050) = 21 • 32 • 52 = 450

Closely related to GCF is the concept of least common multiple(LCM). A somewhat awkward way of finding LCM(30, 96) is towrite lists of multiples of 30 and of 96. Surely 30 • 96 is on both ofthese lists. We are, however, looking for the smallest numbercommon to both lists.

Here we find:

Multiples of 30:30, 60, 90, 120, 150,…, 450, 480, 510,…, 2880, 2910,…

Multiples of 96:96, 192, 288, 384, 480, 776,…, 2880, 2976,…

so that LCM(30, 96) = 480.

However, LCM(30, 96) can also be found as the smallest product ofprimes that contains the prime factorizations of both 30 and 96.Recalling that 30 = 2 • 3 • 5 and that 96 = 2 • 2 • 2 • 2 • 2 • 3 = 25 • 3,we find that

LCM(30, 96) = 25 • 3 • 5 = 480

Have participants find LCM(24, 64) and LCM(32, 48) using primefactorization.

The characterization of GCF and LCM in terms of prime factorizationleads to the following important fact:

GCF(a, b) • LCM(a, b) = a • b.

Having already found that GCF(30, 96) = 6, we could now findLCM(30, 96) as

LCM(30, 96) = 30 x 96

GCF(30, 96)=

30 x 96

6= 5 x 96 = 480 .

Explanation for this formula.

After recalling that GCF(24, 64) = 8, ask participants to findLCM(24, 64) by applying GCF(a, b) • LCM(â, b) using H-15.





T-16

T-16A/H-16

T-16B

T-17

T-18

LCM’s play an important role in the addition of fractions. In order to

add 1

12+

3

8, we have to rewrite this problem in terms of equivalent

fractions with a common denominator.

One possible choice for a common denominator is 8 • 12, leading to8

96+

36

96=

44

96=

11

24.

A more convenient choice is the least common denominator, which is

LCM(12, 8) = 24. This leads to 2

24+

9

24=

11

24.

Have participants add fractions by finding least common denominatoron worksheet H-16.

Practice word problems for LCM and GCF. Cover the answers.

In order to show that a given number is prime, it is necessary to showthat its only factors are 1 and the number itself. For example,

The factors of 113 are: 1, 113

and for that reason 113 is prime.

Enrichment – Determining Whether a Number is PrimeA number that is a perfect square is surely not prime. Recalling thatthe factors of all other non-prime numbers occur in pairs, it is notnecessary to check all smaller numbers as possible factors.

For example, to determine whether 137 is prime, we note that 137 <144, where 144 is a perfect square. For any pair of factors whoseproduct is 137, one of the factors would be less that 137 . Since

121 < 137 < 14411< 137 < 12

it is sufficient to confirm that the numbers 2, 3, 4, 5, 6, 7, 8, 9, 10, and11 are not factors of 137. Having done this, we can conclude that 137is prime.

In fact, it is not even necessary to check all whole numbers between 1and 11. If 6 were a factor of 137, its prime factors (notably 2 and 3)would be factors of 137 as well. On this basis, we need only checkthat the primes 2, 3, 5, 7, and 11 are not factors of 137.





T-19/H-19

T-20T-21T-22T-23

More generally, to establish that a is prime, it is sufficient to confirmthat all primes less than a fail to be factors of a.

Ask participants to determine whether the numbers 223 and 179 areprime. What is the smallest list of numbers that must be checked aspossible factors of 223 and of 179?

Appendix

These slides give more detailed justification for the rules fordetermining divisibility by 2 and 3. The facilitator may use theseslides at his or her discretion. They may be used immediatelyfollowing slide T-6A.

Standards covered in this module:

Grade 4 Number Sense4.0 Students know how to factor small whole numbers.4.1 Understand that many whole numbers break down in

different way.4.2 Know that numbers such as 2, 3, 5, 7, and 11 do not

have factors except 1 and themselves and that suchnumbers are called prime numbers.

Grade 5 Number Sense1.4 Determine the prime factors of all numbers through 50

and write the numbers as the product of their primefactors using exponents to show multiples of a factor.

Grade 6 Number Sense2.4 Determine the least common multiple and greatest

common divisor of whole numbers; use them to solveproblems with fractions.

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers Pre-Post Test


Factors of Whole NumbersPre- Post-Test

1. List all the prime numbers less than 25:

2. Write the prime factorization for each number:

a. 30 = b. 48 =

c. 37 = d. 112 =

3. List all factors of

a) 30

b) 48

c) 37

d) 112

4. What is the greatest common factor of 48 and 112?

5. What is the least common multiple of 128 and 96?

6. State the Fundamental Theorem of Arithmetic.

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers Pre-Post TestAnswer Key


Factors of Whole NumbersPre- Post-Test Answer Key

1. List all the prime numbers less than 25:

2, 3, 5, 7, 11, 13, 17, 19, 23

2. Write the prime factorization for each number:

a. 30 = 2 • 3 • 5 b. 48 = 2 • 2 • 2 • 2 • 3= 24 • 3

c. 37 = 37 d. 112 = 2 • 2 • 2 • 2 • 7= 24 • 7

3. List all factors of

a) 30 1, 2, 3, 5, 6, 10, 15, 30

b) 48 1, 2, 3, 4, 6, 8, 12 16, 24, 48

c) 37 1, 37

d) 112 1, 2, 4, 7, 8, 14, 16, 28, 56, 112

4. What is the greatest common factor of 48 and 112?

16

5. What is the least common multiple of 128 and 96?

384

6. State the Fundamental Theorem of Arithmetic

Every number has a unique prime factorization.

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-1


Divisibility

A whole number a is divisible by another(smaller) counting number b if a = bq where q isa whole number.

In this case we write a ÷ b = q

b is a factor or divisor of a

Example: a = 15 is divisible by b = 3

because 15 = 3q where q = 5

Equivalently, 15 ÷ 3 = 5

Notice that a = bq is a special case of a = bq + rwhen r = 0. So a is divisible by b means that theremainder is zero when a is divided by b.



Zero and One

Any whole number a is divisible by 1.

a ÷ 1 = a because a = 1 • a

0 is divisible by any number a ≠ 0

0 ÷ a = 0 because 0 = a • 0



Be careful!

0 • a = 0

means that you can divide a number into zero

What about dividing a number by zero?

a ÷ 0 does not make sense

suppose a ÷ 0 = qthen a = 0 • q

This is impossible if a ≠ 0.

If a = 0 then any value of q works,but for division there can only be one answer.



Visualization of a = bq

24 = 4 • 6

q = 6• • • • • •

b = 4 • • • • • •• • • • • •• • • • • •

a) Is there another rectangular arrayfor 24?

b) How many arrays are there for 7?

Answer: For 7 there is only one array.

• • • • • • •

7 = 1 • 7

because 7 is a “prime” number

71



A whole number is called prime if it hasexactly two factors,

One and itself.

The list of primes begins:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,……..

There is no largest prime

A number > 1 that is not prime is called

composite



Divisibility Rules

There are well-known shortcuts for testing if wholenumbers are divisible by other whole numbers.

Number Shortcut2: Ones digit is 0 or even

(0, 2, 4, 6, 8)

3: Sum of digits is divisible by 3

4: Last two (tens and ones) digitsare divisible by 4

5: Ones digit is zero or 5

6: Rules for 2 and 3 both work

8: Last 3 digits (hundreds, tens,and ones) are divisible by 8

9: Sum of digits are divisible by 9

10: Ones digit is zero

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-6A/H-6


Practice without calculator

1. Is 324 divisible by 4?



PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-6B


Why does the divisibility rule for 2 work?

354 ÷ 2 = 12

3•102 +5•10 + 4( )

3•102

2+ 5•10

2+ 4

2

Since 10 is divisible by 2, so is 102

Therefore, 3 •102

2= 3•

102

2 and

5 •102

= 5•102

are both whole numbers.

This is true even if 3 in the hundred’s column

and 5 in the ten’s column were replaced by

other numbers.

So, 354 ÷ 2 is a whole number exactly when

the number in the one’s column is divisible by

2. The number in the one’s column must

be 0, 2, 4, 6 or 8.

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-6C


Why does the divisibility by 3 rule work?

Example: Is 651 divisible by 3?

Yes, because 6 + 5 + 1 = 12 is divisible by3. Why does this work?

651 ÷ 3 = 13

6•102 + 5•10 +1( )

= 13

6(99 +1)+ 5(9 +1)+1[ ]

= 13

6 • 99 +6 + 5• 9 + 5 +1[ ]

= 13

(6 • 99 + 5• 9) + 6 + 5 +1[ ]

= 6• 99

3+ 5•9

3+ 6 + 5 +1

3

Since any multiple of 99 and any multipleof 9 are divisible by 3, 651 is divisible by 3if and only if 6 + 5 + 1 is divisible by 3.



Every composite number > 1 can be written as theproduct of primes.For example:

“Chipaway” technique for 2424 = 2 • 12

= 2 • 2 • 6= 2 • 2 • 2 • 3= 23 • 3

“Split Asunder” technique for 2424 = 4 • 6

= 2 • 2 • 2 • 3= 23 • 3

“Chipaway” technique for 12001200 = 2 • 600

= 2 • 2 • 300= 2 • 2 • 2 • 150= 2 • 2 • 2 • 2 • 75= 2 • 2 • 2 • 2 • 3 • 25= 2 • 2 • 2 • 2 • 3 • 5 • 5= 24 • 3 • 52

“Split Asunder” technique for 12001200 = 30 • 40

= 5 • 6 • 5 • 8= 5 • 2 • 3 • 5 • 2 • 2 • 2= 24 • 3 • 52



Factor trees can be useful to arrive at primefactorizations.

Using the “Chipaway” technique for 24

24

2 • 12

2 • 2 • 6

2 • 2 • 2 • 3

leading to:

24 = 2 • 2 • 2 • 3

= 23 • 3

Using the “Split Asunder” technique for 24

24

4 • 6

2 • 2 • 2 • 3

leading to:

24 = 2 • 2 • 2 • 3

= 23 • 3



Factors of Whole Numbers - Worksheet

Factor 60 and 500 by drawing factor trees forboth techniques.

“Chipaway” Technique:

60 = 500 =

“Split Asunder” Technique:

60 = 500 =



Fundamental Theorem of Arithmetic:

Every composite number greater than one can

be expressed as a product of prime numbers.

Except for the order in which the prime

numbers are written, this can only be done in

one way.

Except for order, you will obtain the same list

of primes regardless of the method used to

arrive at a prime factorization.

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-10A


Factors

Factors of 24

24 = 1 • 242 • 123 • 84 • 6

So the list of all factors of 24 is

1, 2, 3, 4, 6, 8, 12, 24

Factors of 37

37 = 1 • 37 = 37 • 1

So the list of all the factors is

1, 37

Factors of 64

64 = 1 • 642 • 324 • 168 • 8

So the list of factors of 64 is

1, 2, 4, 8, 16, 32, 64

(Note 8 is only listed once)



Factors

Factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.

Factors of 37 are: 1, 37.

Factors of 64 are: 1, 2, 4, 8, 16, 32, 64.

Except for perfect squares, factors appear in

pairs whose product is the number being

factored.

1, 2, 3, 4, 6, 8, 12, 24

1, 37

1, 2, 4, 8, 16, 32, 64

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-11/H-11


Practice

Find the prime factors of 32 and 80.

Find all factors of 32 and 80.



Common Factors

Recalling the following lists of factors

24: 1, 2, 3, 4, 6, 8, 12, 24

64: 1, 2, 4, 8, 16, 32, 64

We see that the common factors of 24 and 64are:

1, 2, 4 and 8

This leads to the greatest common factor(GCF):

The GCF of 24 and 64 is 8

or GCF(24, 64) = 8

Practice:

a) Find the GCF(30, 96)

b) Find the GCF(90, 75)



GCF by Prime Factorization

It is also possible to determine GCF(a,b)from each number’s prime factorization.

Example: Find GCF (360, 270)

360 = 23 • 32 • 51

270 = 21 • 33 • 51

Choose the smallest power of each primethat occurs in either list.

GCF(360, 270) = 21 • 32 • 51 = 90

Question: How does this work if there aredifferent primes for each number?

Answer: Use the zero exponent.



GCF and the Zero Exponent

Example: Find GCF(84, 90)

84 = 22 • 3 • 7

90 = 2 • 32 • 5

Rewrite so that all primes appear in bothlists. Use the zero exponent.

84 = 22 • 31 • 50 • 71

90 = 21 • 32 • 51 • 70

Now choose each prime raised to thelowest power.

GCF(84, 90) = 21 • 31 • 50 • 70

= 2 • 3 • 1 • 1

= 6

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-13C/H-13


GCF Practice

Find the GCF of 30 and 96 by both methods.

Find the GCF of 90 and 75 by both methods.

Find the GCF of 4500 and 4050 by primefactorization.



Least Common Multiple

One way to find LCM(30, 96) is to write lists of

multiples of 30 and of 96.

The least common multiple is the smallest number

common to both lists of multiples.

Multiples of 30:

30, 60, 90, 120, 150,…450, 480,

510,…,2880, 2910,…

Multiples of 96:

96, 192, 288, 384, 480, 776,…,2880, 2976,…

From these lists we see that LCM(30, 96) = 480.



However, LCM(30, 96) can also be found

using prime factorizations.

30 = 22 • 31 • 51

96 = 25 • 31 • 50

Notice that 50 = 1, so the prime factorization of

96 is really 25 • 3.

To find LCM’s, make sure that each prime

occurs in both lists.

To find the LCM, choose each prime raised to

the highest power from either list.

LCM(30, 96) = 25 • 31 • 51

= 480



Least Common Multiple

Worksheet

Use prime factorization to find

LCM(24, 64)

LCM(32, 48)



The characterization of GCF and LCM interms of prime factorization leads to:

GCF(a, b) • LCM(a, b) = a • b .

Example: GCF(30, 96) • LCM(30, 96) = 30 • 96

GCF(30, 96) = 6

LCM(30, 96) = 480

6 • 480 = 30 • 96

2880 = 2880

Using GCF(30, 96) = 6, we could have foundLCM (30, 96) as

LCM(30,96) =30 • 96

GCF(30, 96)= 30•96

6

= 5 • 96

= 480

LCM(a,b) =a • b

GCF(a, b)



Why does

GCF(a,b ) • LCM(a,b ) = a • b ?

Consider an example:

a = 4500 = 22 • 32 • 53

b = 4050 = 21 • 34 • 52

GCF(a, b ) = 21 • 32 • 52

(Choose the smallest powers)

LCM(a, b ) = 22 • 34 • 53

(Choose the largest powers)

But a • b = 22 • 32 • 53 • 21 • 34 • 52,

and this is the same as

GCF(a , b ) • LCM (a , b )

Both are 18,225,000



Least Common MultiplesWorksheet

Find LCM(24, 64) by applying

GCF(a, b) • LCM(a, b) = a • b.

Find LCM(32, 48) by applying

GCF(a, b) • LCM(a, b) = a • b



LEAST COMMON MULTIPLE ANDADDITION OF FRACTIONS

Least common multiples help with theaddition of fractions.

For example1

12+

38

= ?

We rewrite this problem in terms ofequivalent fractions with a commondenominator.

One possible choice for a commondenominator is 8 • 12,

896

+3696

=4496

=1124

Or,

by using LCM(12, 8) = 24

224

+9

24=

1124



Addition of FractionsWorksheet

Find the sums for the following fractionsby using the least common denominator.

1.23

+ 45

= 2.13

+ 512

=

3.2

34+ 5

24= 4.

130

+ 396

=

5.5

16+ 3

56= 6.

251

+ 568

=



Sample Word Problems

Try these problems:

1. Two joggers are running around a track.One jogger runs a lap in 6 minutes andthe other takes 10 minutes. If they start atthe same time and place, how long will ittake for them to meet at the starting placeat the same time.

2. A 48 member band follows a 54 memberband in a parade. If the same number ofplayers must be in each row of bothbands, what is the largest number ofplayers that can be placed in each row?Assume each row has the same number ofplayers.

Answers:1. LCM(6, 10) = 302. GCF(48, 54) = 6



EnrichmentDetermining Whether a Number is Prime

For a number that is not a perfect square,factors occur in pairs. To determine whether anumber is prime, it is not necessary to checkall smaller numbers as possible factors.

Is 137 prime?

For any pair of factors whose product is 137,one of the factors would be less than 137.

121< 137< 144

11 < 137 < 12

It is sufficient to confirm that the numbers

2, 3, 4, 5, 6, 7, 8, 9, 10, and 11

are not factors of 137

to show that 137 is prime.



In fact, it is not necessary to check all whole

numbers between 1 and 11.

E.g., if 6 were a factor of 137,

its prime factors (2 and 3)

would also be factors of 137.

We only need to check that the primes < 137

2, 3, 5, 7, and 11

are not factors of 137.

To establish that a is prime,

it is sufficient to confirm that

all primes less than or equal to a

fail to be factors of a.

PRIMARY CONTENT MODULE IV NUMBER SENSE: Factors of Whole Numbers T-19/H-19


Checking Whether a Number is Prime

Worksheet

What is the smallest list of numbers that must bechecked as possible factors of 223?

Is 223 prime?

What is the smallest list of numbers that must bechecked as possible factors of 179?

Is 179 prime?



Divisibility Rules

Basic ingredients:

“A|B” means “A divides B or in other words,

there is an integer k such that

B = A • k.

Example: 2|10 because 10 = 2 • 5

Theorem: If A|B and A|C , then A|(B+C).

Example: If 2|10 and 2|8, then 2|18

Proof: If A|B , there is an integer k1, such that

B = A • k1. If A|C, there is an integer

k2 such that C = A • k2.

Therefore: B + C = A • k1 + A • k2

= A (k1 + k2)

So, A|(B + C)



Here is another result

Theorem 2: If A|(B + C) and A|C, then A|B

Example: If 2|18 and 2|10, then 2|(18-10) or

2|8.

Proof: If A|(B+C), B + C = A • k1 , for some

integer k1 . If A|C, C = A • k2 for some integer

k2. Then

B = (B + C) – C = A • k1 – A • k2

= A (k1 – k2)

Therefore A|B.



Divisibility by 2

2|(a •102 + b • 10 + c ) if and only if 2|c.

Proof: Let B = a• 102 + b • 10. Then 2|B

because B = 2 • (A • 50 + 6 • 5). If 2|c , then

2|(B + c) by Theorem 1. If 2|(B + c ), then 2|c

by Theorem 2.

The only single digit numbers divisible by 2

are 0, 2, 4, 6, and 8.

Therefore, 2|(a • 102 + b • 10 + c) if and only

if c = 0, 2, 4, 6, or 8.



Divisibility by 3

3|(a • 102 + b • 10 + c ) if and only if

3|(a + b + c )

Proof:

a • 102 +b • 10 + c = a • (99 + 1) + b • (9 + 1) + c

= (a • 99 + b • 9) + (a + b + c)

Let B = a • 99 + b • 9. Then 3|B

Because B = 3 • (a • 33 + b • 3).

Let C = a + b + c . If 3|C , then 3|(B + C) by

Theorem 1. If 3|(B +C ) then 3|C by Theorem

2.

Example: 3|129 because 3|(1+2+9)

Documents

The Winning EQUATION - California State University, …vcmth00m/factors.pdfdivide any whole number a by any counting number b, assuming a > b, special importance is attached to the